Peltier effect in psychological study

[Lewishodgson]

Hi, I am a neuroscientist developing a study on pain responses. For this purpose I wish to use a Peltier element to produce a controlled level of heat. I have purchased a 3x3cm elements rated between 5v - 16v max with 3.3 amp max current. I wish to be able to control heat output. In the lab we have a variable DC current source. What I need to know is whether any additional control elements will need to be introduced into the circuit to produce the desired effect? This is a methodology that has been used previously in psychology studies so I know it must work but have been unable to find any constructional details in the published literature. I would be most grateful for any assistance anyone can provide. Many thanks.

 

[colin55]

Place a 1 ohm 5watt wire-wound resistor in series with the element and increase the voltage until you get the desired heat/cooling effect.
Note: the current must not be greater than 3 amp.

 

[Lewishodgson]

Peltier reply

Hi Colin, many thanks for your superfast response. I will try this right away. David

 

[ke5frf]

Hi, I am a neuroscientist developing a study on pain responses. For this purpose I wish to use a Peltier element to produce a controlled level of heat. I have purchased a 3x3cm elements rated between 5v - 16v max with 3.3 amp max current. I wish to be able to control heat output. In the lab we have a variable DC current source. What I need to know is whether any additional control elements will need to be introduced into the circuit to produce the desired effect? This is a methodology that has been used previously in psychology studies so I know it must work but have been unable to find any constructional details in the published literature. I would be most grateful for any assistance anyone can provide. Many thanks.

Colin's response: "Place a 1 ohm 5watt wire-wound resistor in series with the element and increase the voltage until you get the desired heat/cooling effect.
Note: the current must not be greater than 3 amp. "

IS there some reason I am unaware of for adding a series resistor here? All you will be doing is dropping the voltage across the resistor and wasting power.

Peltiers are load devices which draw their own current. As long as the supplied voltage doesn't exceed what the rating indicates it won't exceed more than 3 amps and all will be well.

My advice to the OP would be to build a feedback temperature controller or purchase one. If building such a circuit is more than you have time or expertise to do, I suggest purchasing one from a lab supplier like Watlow, Omega, Fisher Scientific, etc. That is if the project is funded.

If your power supply is variable then you should be able to manually control the temperature easily enough by adjusting the voltage within the specified range.

You said something about a "variable current source". So that we don't confuse terminology, is this a variable DC power supply or an actual regulated current source? If you are actually manipulating the current instead of the voltage, by nature of Ohm's Law the proper voltage would be attained when the current indication is less than 3 amps, you can prove this with a DC voltmeter across the load. At 3 amps you should be no more than 15 to 16 volts.

But again, the most optimal set-up would be either an "on/off" thermostatic or analog PID controller with a sensing element providing feedback to maintain automated temperature.

And if you choose to go a manual route, the resistor is uneccessary. You have variable control with your power supply. If anything I would add a series 2.5 to 3 amp fuse instead of a resistor to ensure the rated current is never exceeded by an accidental "overvoltage".

With manual control, you could easily monitor the surface temperature of the test specimen with an inexpensive digital thermometer purchased at a hardware store, electrical supply store, or even perhaps a department store. A cheap thermometer designed for checking food in the oven could be hacked and extra lead length added to the sensing element (thermistor, rtd, ic chip)

This would be a very easy thing to do.

 

[colin55]

I put a safety resistor in place because no Peltier device 3cm x 3cm is going to allow a dissipation of 45 watts.
Something is wrong somewhere and I put a margin of safety in place so that the device is not damaged.

 

[ke5frf]

Another point. I did some quick math with approximate numbers and roughly figured the power of the peltier at 3 amps and rated voltage to be 48 watts...this is given the figure of "16V max" as the supply voltage that will produce the 3 amps.

Roughly, a 1 ohm resistor will drop 8 watts at this condition. A 5 watt resistor will smoke.
Not a good idea, IMHO. My approximated math may be wrong though so verify it for yourself.

 

[ke5frf]

I put a safety resistor in place because no Peltier device 3cm x 3cm is going to allow a dissipation of 45 watts.
Something is wrong somewhere and I put a margin of safety in place so that the device is not damaged.

It will if properly sinked. This is what they are designed to do. I do not imagaine the full 45 (or 48 or whatever) watts will be used if this is making human skin contact. A fuse would be a superior protection device.

 

[colin55]

ke5frf How can a 1 ohm resistor with 3 amps flowing, dissipate 8 watts.
You should really not be responding to these posts as you know nothing about electricity, let alone electronics.
.

 

[ke5frf]

ke5frf How can a 1 ohm resistor with 3 amps flowing, dissipate 8 watts.
You should really not be responding to these posts as you know nothing about electricity, let alone electronics.
.

3 amps x 16 volts= 48 watts.

R=E/I

R=16v/3 amps

R=5.333 (round to 5 for ease)

5+1=6 ohms

5/6 x 48= 40 watts across peltier

8 watts across 5 watt resistor.

 

[ke5frf]

It won't dissipate 8 watts because it will be burnt in two.

 

[ke5frf]

All it takes is less than 2 volts to be dropped across a 1 ohm resistor at 3 amps to create more than 5 watts and destroy it. This is the most basic electronics math there can be. FYI.

 

[colin55]

ke5frf Do not make any more posts as you know absolutley nothing about electrical circuits.

 

[ke5frf]

ke5frf Do not make any more posts as you know absolutley nothing about electrical circuits.

Then please show the error in my math.

 

[colin55]

I must be going mad.

 

[ke5frf]

Ummm, no.

Use your own formula correctly.

P=I x I x R

I=current

current = 3 amps

P = 3 x 3 x 1 (R is ohms, ohms=1)

P = 9 watts.

The 1 watt I lost in my approximate math was removed when I rounded the .333 off of 5.333 ohms

Please quit while you are ahead my friend. I am trying to help this gentleman build a safe piece of equipment that won't cause injury or damage.

 

[colin55]

You are correct. I must be going mad.

 

[ke5frf]

You are correct. I must be going mad.

That is OK, as long as the proper answer was reached who cares.

 

[kinarfi]

let's start with 10 volts and 1.5 amps, 15 watts, applied to a gallon of water for 1 minute, nothing, applied to skin via 3 cm square piece of aluminum for 1 minute, probably some very severe burns and blisters, you better get yourself a very accurate temperature controller or more insurance. A good temp controller will let you know if you giving first, second or third degree burns to your victims, what are you a sadist?
kinarfi

 

[ke5frf]

To be fair, we do not know if the OP is directly applying the heat, applying it through a sink, or really any detail. But your warnings are certainly warranted.

This is why I recommended a temperature controller very early on before the thread was thrown off track. At the very minimum the temperature of the application should be directly monitored, even if control is manually administered.

 

[Roff]

All it takes is less than 2 volts to be dropped across a 1 ohm resistor at 3 amps to create more than 5 watts and destroy it. This is the most basic electronics math there can be. FYI.

How do you get less than 2V across a 1 ohm resistor with 3 amps flowing through it?
V=I*R=3*1=3V